Below is a theorem that is equivalent to Brouwer fixed-point theorem, which I found quite interesting. The proof is in this PDF file.

Let $v: \mathbb S^{n-1} \to \mathbb R^n$ be a continuous map, satisfying $v(x) \cdot x \leq 0$. Then we can't extend it to be a continous map $\tilde v: B^n \to \mathbb R^n$ such that $\tilde v(x) \neq 0$ for every $x \in B^n$.

The proof of above theorem goes like this. Suppose such $\tilde v$ exists, then we can define a map $u: B^n\to S^{n-1}$ such that $u(x) = \tilde v(x)/|\tilde v(x)|$. From Brouwer fixed-point theorem, there is a $x_0 \in B^n$ satisfying $u(x_0)=x_0$, which implies $x_0\in \mathbb S^{n-1}$. However, $u(x_0) \cdot x_0 = x_0 \cdot x_0 > 0$, which contradicts to the condition $v(x_0) \cdot x_0 \leq 0$.

My question is, given a nonzero map $v: \mathbb S^{n-1} \to \mathbb R^n$, is there a criterion to decide whether we can extend it to a nonzero map $\tilde v: B^n \to \mathbb R^n$?

For example, given a constant nonzero field $v(x) = v_0$, we can definitely extend it to be the same constant vector on $B^n$.

Background: When I read an ODE textbook, I encountered a corollary of Poincaré-Bendixson theorem, which is similar to the theorem above. So I found the above theorem when trying to relate the corollary to Brouwer fixed-point theorem. For me it is very natural to ask the question: under what condition can we extend a nonzero vector field on sphere to a nonzero vector field on ball. I think it can help gain more insight into topology of vector fields.

There are also some related posts from MSE: the-hairy-ball-theorem-from-brouwers-fixed-point and hairy-disk-theorem.

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    $\begingroup$ $v:S^{n-1}\to\mathbb{R}^n\setminus \{0\}$ defines a continuous map $\bar{v}=\frac{1}{\Vert v\Vert}v: S^{n-1}\to S^{n-1}$. $v$ extends iff $\deg\bar{v}=0$. $\endgroup$ Feb 12 at 9:52

1 Answer 1


View $B^{n}$ as $(S^{n-1}\times[0,1])/(S^{n-1}\times \{1\})$. This means that an extension of a map from $B^{n}$ to somewhere is just an extension to $S^{n-1}\times [0,1]$ which is constant on the other end. Such a map is usually called a nullhomotopy.

That somewhere is in your question $\mathbb{R}^n\setminus\{0\}$, which is homotopy equivalent to $S^{n-1}$. So you are now asking for the set of all homotopy classes from $S^{n-1}\to S^{n-1}$. For this specific space (and for $n\ge 2$), these are given by the degree of a map, which generalizes the notion of a winding number. The map is extensible, if and only if the degree is zero.

The winding number associated to any map $S^1\to S^1$ the number of times that it walks around the circle. It can be defined by first making the map smooth, then picking a point in the range and counting how many times the map crosses that point preserving the orientation and subtracting how many times it crosses that point reversing the orientation.

And that is also the definition of the degree of a map (in higher dimensions).

Note that for $n=2$ the condition on the scalar product from this question forces the winding number to be one. This can be visualized as follows. Pick one point $x_0$ in the circle and mark the allowed points red. Then visualize what happens when you move $x_0$ once around the circle.

EDIT: I corrected the dimensions as Didier suggested.

  • $\begingroup$ Thank you for the very clear explnation! I have yet another question: when $n>2$, what can we say about the degree of map from the scalar product condition? Are there explicit formulas for calculation of the degree of map? I think it would be nice if we can rephrase the proof in question by the notion of degree of map. $\endgroup$ Feb 12 at 10:25
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    $\begingroup$ You can write down an explicity homotopy of $v$ to $-id$. Just connect $v(x_0)$ and $-sx_0$ by a straight line in $\mathbb{R}^n$. This line does not go through the origin, since $v(x_0)\neq x_0$ by the given condition. Then project that line to $S^{n-1}$. Thus it remains to determine the degree of $-id$, which is $\pm 1$, depending on whether it is orientation preserving or reversing. This also shows that the scalar product condition can be weakened further. $\endgroup$ Feb 12 at 10:33
  • $\begingroup$ Excuse me, shouldn't it be $B^n=(S^n\times [0, 1])/(S^n\times \{0, 1\})$? $\endgroup$ Feb 12 at 13:54
  • $\begingroup$ No that is a rather weird space. Collapsing the top and a bottom to one point each would give $\mathbb{S}^{n+1}$, and not $B^n$. Identifying the points with each other gives a space homotopy equivalent to $\mathbb{S}^{n+1}\vee \mathbb{S}^1$. $\endgroup$ Feb 12 at 16:57
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    $\begingroup$ The dimensions do not match. I think you meant $B^{n+1} = S^n\times[0,1] / S^n\times \{1\}$, if we think of the same thing: the cone over $S^n$, that is, the cylinder over $S^n$ with one boundary component collapsed into a point. $\endgroup$
    – Didier
    Feb 12 at 19:47

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